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Question Number 85456 by jagoll last updated on 22/Mar/20

$$\int \sqrt{\csc \mathrm{x}}\mathrm{dx}?$$

Commented by jagoll last updated on 22/Mar/20

$$\mathrm{thank}\mathrm{sir}$$

Commented by MJS last updated on 22/Mar/20

$$\mathrm{sadly}\mathrm{this}\mathrm{is}\mathrm{wrong}.$$$$t=\sqrt{\sin x}\to dx=\frac{2\sqrt{\sin x}}{\cos x}dt=\frac{2t}{\sqrt{1-t^4}}dt$$$$\Rightarrow 2\int \frac{dt}{\sqrt{1-t^4}}\mathrm{and}\mathrm{we}{\mathrm{can}}^{\prime }\mathrm{t}\mathrm{solve}\mathrm{this}$$

Commented by MJS last updated on 22/Mar/20

$$\mathrm{you}\mathrm{should}always\mathrm{test}\mathrm{your}\mathrm{solutions}$$$$\frac{d}{dx}\left[2\arcsin \sqrt{\sin x}\right]=\frac{\cos x}{\sqrt{\sin x-{\sin }^{2}x}}\ne \frac{1}{\sqrt{\sin x}}$$

Commented by jagoll last updated on 22/Mar/20

$$\mathrm{so}\mathrm{what}\mathrm{the}\mathrm{correct}\mathrm{answer}\mathrm{sir}?$$

Commented by MJS last updated on 22/Mar/20

$$\mathrm{I}\mathrm{cannot}\mathrm{solve}\mathrm{it}$$$${\mathrm{there}}^{\prime }\mathrm{s}\mathrm{a}\mathrm{solution}\mathrm{using}\mathrm{\Gamma }\left(x\right)\mathrm{somehow},\mathrm{it}\mathrm{had}$$$$\mathrm{been}\mathrm{on}\mathrm{this}\mathrm{forum},\mathrm{maybe}\mathrm{a}\mathrm{year}\mathrm{ago},\mathrm{I}{\mathrm{can}}^{\prime }\mathrm{t}$$$$\mathrm{find}\mathrm{it}$$

Commented by jagoll last updated on 22/Mar/20

$$\mathrm{thank}\mathrm{you}\mathrm{sir}$$

Commented by MJS last updated on 22/Mar/20

$$\mathrm{found}\mathrm{this}:$$$$\int \frac{dx}{\sqrt{\sin x}}=$$$$[t=x-\frac{\pi }{2}\to dx=dt]$$$$\int \frac{dt}{\sqrt{\cos t}}=\int \frac{dt}{\sqrt{1-{2\sin }^2\frac{t}{2}}}=$$$$=\mathrm{F}(\frac{t}{2}\mid 2)=\mathrm{F}(\frac{x}{2}-\frac{\pi }{4}\mid 2)+C$$$${\mathrm{it}}^{\prime }\mathrm{s}\mathrm{an}\mathrm{eliptic}\mathrm{integral}$$$$\mathrm{but}\mathrm{I}\mathrm{cannot}\mathrm{explain},{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{not}\mathrm{my}\mathrm{solution}$$

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